Light 1289 LCM from 1 to n 素数筛选位优化

题目来源：Light 1289 LCM from 1 to n

题意：。。

思路：从1到n 打过某个数是以一个素数的几次方 那么答案就乘以这个素数

主要是筛选素数 存不下 位优化 一个整数32位标记32个数 内存缩小32倍

是学习别人的

#include <cstdio>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
const int maxn = 100000010;
const int maxm = 6000000;
unsigned int dp[maxm];
int prime[maxm];
int vis[maxn/32+10];
//筛素数 
int sieve()
{
	//memset(vis, 0, sizeof(vis));
	//vis[0] = vis[1] = 1;
	prime[0] = 2;
	dp[0] = 2;
	int c = 0;
	for(int i = 3; i < maxn; i += 2)
	{
		if(!(vis[i/32]&(1<<(i%32))))
		{
			prime[++c] = i;
			dp[c] = dp[c-1] * i;
			for(int j = i*2; j < maxn; j += i)
				vis[j/32] |= (1<<(j%32));
		}
	}
	return c;
}

int main()
{
	int c = sieve();
	int cas = 1;
	int T;
	scanf("%d", &T);
	while(T--)
	{
		int n;
		scanf("%d", &n);
		int l = 0, r = c-1, m;
		while(l <= r)
		{
			int mid = (l + r) >> 1;
			if(prime[mid] <= n)
			{
				m = mid;
				l = mid + 1;
			}
			else
				r = mid - 1;
		}
		//printf("%d\n", m);
		unsigned int ans = dp[m];
		for(int i = 0; i <= m && prime[i]*prime[i] <= n; i++)
		{
			int x = prime[i];
			int y = prime[i]*prime[i];
			
			while(y <= n && y / x == prime[i])
			{
				//printf("**%d", y);
				ans *= prime[i];
				x *= prime[i];
				y *= prime[i];
			}
			//ans *= x;
		}
		printf("Case %d: %u\n", cas++, ans);
	}
	return 0;
}

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Light 1289 LCM from 1 to n 素数筛选位优化

原文：http://blog.csdn.net/u011686226/article/details/29843851